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I want to generate a combination of vectors of $N$ dimensions so that their directions are as different as possible if you go down the list.

So, if $N=2$, the first would be $[1, 0]$, the second would be $[-1, 0]$, the third would be $[0, 1]$, the fourth would be $[0, -1]$, and so forth, but extend this to allow for any number of dimensions.

No table lookup, please. I need an algorithm that can do this.

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Don't you just want a list of vectors that have 1 or -1 is some position and 0 everywhere else? Or can there be more vectors that (twice the) dimensions? – Karolis Juodelė Oct 3 '12 at 16:57
There can be more vectors than dimensions. It basically needs to be a complete combination, but sorted in a way where the difference in direction between the vectors as an aggregation will always be maximized while going through the list top to bottom. – user644337 Oct 3 '12 at 17:01
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Well, the first and the third have the same direction ;-) And the "so forth" in the example leaves me wondering... does the sequence (for N=2) continues? – leonbloy Oct 3 '12 at 17:07
The first and third are not the same. One points down and the other points right. The 5th one in the sequence could be [1, 1], [-1, -1], [1, -1], or [-1, 1] – user644337 Oct 3 '12 at 21:56

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1 Answer

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What about these $2n$ vectors: $(\pm 1,0,0,\dots,0)$, $(0,\pm 1,0,\dots,0)$, $(0,0,\pm 1,\dots,0)$, ..., $(0,0,0,\dots,\pm 1)$?

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What is the number of vectors as a function of $N$? – Stefan Smith Oct 3 '12 at 22:30

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